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01 May 2012, 08:29

Bunel,

I have read your post on this topic in detail. But the reason why the inequality sign was flipped (from the original question) for the first part of the expression (1-2x)>0 and not for the second part of the expression (1 + x)<0 is not clear.

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I have read your post on this topic in detail. But the reason why the inequality sign was flipped (from the original question) for the first part of the expression (1-2x)>0 and not for the second part of the expression (1 + x)<0 is not clear.

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This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If x is an integer, is |x| > 1?

First of all: is \(|x| > 1\) means is \(x<-1\) (-2, -3, -4, ...) or \(x>1\) (2, 3, 4, ...), so for YES answer \(x\) can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as \((2x-1)(x+1)>0\) (so that the coefficient of x^2 to be positive after expanding): roots are \(x=-1\) and \(x=\frac{1}{2}\) --> "\(>\)" sign means that the given inequality holds true for: \(x<-1\) and \(x>\frac{1}{2}\). \(x\) could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as \((x-1)(2x+1)>0\): roots are \(x=-\frac{1}{2}\) and \(x=1\) --> "\(>\)" sign means that the given inequality holds true for: \(x<-\frac{1}{2}\) and \(x>1\). \(x\) could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is \(x<-1\) and \(x>1\). Sufficient.

Show Tags

This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If x is an integer, is |x| > 1?

First of all: is \(|x| > 1\) means is \(x<-1\) (-2, -3, -4, ...) or \(x>1\) (2, 3, 4, ...), so for YES answer \(x\) can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as \((2x-1)(x+1)>0\) (so that the coefficient of x^2 to be positive after expanding): roots are \(x=-1\) and \(x=\frac{1}{2}\) --> "\(>\)" sign means that the given inequality holds true for: \(x<-1\) and \(x>\frac{1}{2}\). \(x\) could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as \((x-1)(2x+1)>0\): roots are \(x=-\frac{1}{2}\) and \(x=1\) --> "\(>\)" sign means that the given inequality holds true for: \(x<-\frac{1}{2}\) and \(x>1\).\(x\) could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is \(x<-1\) and \(x>1\). Sufficient.

Answer: C.

Hi Bunuel,

I understood the question and got the answer right by finding the intersection for inequalities. The approach I normally use for such questions is as follows:

(1 - 2x)(1 + x) < 0

for above statement to be true the 2 expressions should have opposite signs, giving us (x > 1/2 and x > -1) OR (x < 1/2 and x < -1)Now, finding intersection gives us x > 1/2 OR x < -1. As a result, x < -1 is fine but x > 1/2 includes 1 as an option - making the statement insufficient.

Now my question is, how did you use the logic below to skip some of the steps above and directly go to the conclusion: x<-1 and x> 1/2. . What's the rule if instead of greater than (>) sign it was less than (<).

If I can understand the logic, I believe it will help save some valuable time.